Part 12 - State feedback | rezarezvan.com

In this part we’ll see how we can represent a system using its states.

We can represent a system using its state-space representation.

Given an input signal, $r(t)$ , that we multiply with a constant, $K_r$ .

We also have noise in the system.

Our feedback is also multiplied with a matrix $L_u$ .

x = \begin{bmatrix} x_1 \newline x_2 \newline \vdots \newline x_n \end{bmatrix}

We can represent our system as:

\begin{cases} \dot{x} & = Ax + Bu + B_v \cdot v \newline y & = Cx. \end{cases}

We can define $u(t)$ as:

u = K_r r - L_u x.

Let’s find the transfer function $G_{ry}$ , so we set the noise to zero.

\begin{cases} \dot{x} & = Ax + Bu \newline y & = Cx \end{cases}

\begin{cases} \dot{x} & = Ax + B(K_r r - L_u x) \newline y & = Cx \end{cases}

\begin{cases} \dot{x} & = (A - BL_u)x + BK_r r \newline y & = Cx \end{cases}

If we take the Laplace transform of the state-space representation:

\begin{cases} sX(s) & = (A - BL_u)X(s) + BK_r R(s) \newline Y(s) & = CX(s) \end{cases}

\begin{cases} sX(s) - (A - BL_u)X(s) & = BK_r R(s) \newline Y(s) & = CX(s) \end{cases}

\begin{cases} (sI - A + BL_u)X(s) & = BK_r R(s) \newline Y(s) & = CX(s) \end{cases}

\begin{cases} X(s) & = (sI - A + BL_u)^{-1}BK_r R(s) \newline Y(s) & = CX(s) \end{cases}

\begin{cases} X(s) & = (sI - A + BL_u)^{-1}BK_r R(s) \newline Y(s) & = C(sI - A + BL_u)^{-1}BK_r R(s) \end{cases}

Which means:

G_{ry}(s) = \dfrac{Y(s)}{R(s)} = C(sI - A + BL_u)^{-1}BK_r.

The characteristic equation for this is $\det(sI - A + BL_u) = 0$

We won’t cover the calculations to prove this, but:

L(s) = L_u(sI - A)^{-1} B.

Which means:

\begin{align*} S(s) & = \dfrac{1}{1 + L(s)} = \dfrac{1}{1 + L_u(sI - A)^{-1} B}, \newline T(s) & = \dfrac{L(s)}{1 + L(s)} = \dfrac{L_u(sI - A)^{-1} B}{1 + L_u(sI - A)^{-1} B}. \end{align*}

Let’s now go back, let’s find $G_{ru}$ this time:

\begin{align*} U(s) & = K_rR(s) - L_uX(s) \newline & = K_rR(s) - L_u(sI - A + BL_u)^{-1} BK_r R(s), \newline G_{ru}(s) & = \dfrac{U(s)}{R(s)} = K_r - L_u(sI - A + BL_u)^{-1} BK_r. \end{align*}

Finally, let’s also find $G_{vy}$ , therefore we set $r = 0$ .

\begin{cases} \dot{x} & = Ax - BL_ux + B_v v \newline y & = Cx \end{cases}

\begin{cases} \dot{x} & = (A - BL_u)x + B_v v \newline y & = Cx \end{cases}

Taking the Laplace transform:

\begin{cases} sX(s) & = (A - BL_u)X(s) + B_v V(s) \newline Y(s) & = CX(s) \end{cases}

\begin{cases} (sI - A + BL_u)X(s) & = B_v V(s) \newline Y(s) & = CX(s) \end{cases}

\begin{cases} X(s) & = (sI - A + BL_u)^{-1} B_v V(s) \newline Y(s) & = CX(s) \end{cases}

\begin{cases} X(s) & = (sI - A + BL_u)^{-1} B_v V(s) \newline Y(s) & = C(sI - A + BL_u)^{-1} B_v V(s) \end{cases}

G_{vy}(s) = \dfrac{Y(s)}{V(s)} = C(sI - A + BL_u)^{-1} B_v.

Usually, $K_r$ is found using the fact that we want low-frequency amplification, meaning $G_{ry}(0) = 1$

This means that:

K_r = \dfrac{1}{C(-A + BL_u)^{-1}B}.

Definition 1 (Controllability)

State feedback requires that our control signal, $u(t)$ affects all states. This means that:

\det(S) \neq 0.

Where $S$ :

S = \begin{bmatrix} B & AB & A^2B & \cdots & A^{n-1}B \end{bmatrix} \quad \text{(For clarification, this is a long series of matrix multiplications)}.

We call $S$ for the controllability matrix. The system is controllable if $\det(S) \neq 0$ .

Example 1

We have the following system which has feedback such that:

G_{ry}(s) = \dfrac{8}{s^2 + 4s + 8}.

The given dynamics for the system is represented as:

\begin{cases} \dot{x}_1 & = -2x_1 + x_2 \newline \dot{x}_2 & = -5x_1 + 2u \newline y & = x_1. \end{cases}

We have no noise present in this system.

Let’s firstly determine if this system is controllable.

Let’s define all matrices:

x = \begin{bmatrix} x_1 \newline x_2 \end{bmatrix}

A = \begin{bmatrix} -2 & 1 \newline -5 & 0 \end{bmatrix}

B = \begin{bmatrix} 0 \newline 2 \end{bmatrix}

C = \begin{bmatrix} 1 & 0 \end{bmatrix}

D = \begin{bmatrix} 0 \end{bmatrix}

The controllability matrix:

S = \begin{bmatrix} B & AB \end{bmatrix}

AB = \begin{bmatrix} 2 \newline 0 \end{bmatrix}

S = \begin{bmatrix} 0 & 2 \newline 2 & 0 \end{bmatrix}

\det(S) = -4 \neq 0.

Therefore, this system is controllable.

Let us now find $L_u$ and $K_r$ such that $G_{ry}(0) = 1$ .

Let’s first find $L_u$ .

L_u = \begin{bmatrix} l_1 & l_2 \end{bmatrix}

C(sI - A + BL_u)^{-1}B K_r

\begin{bmatrix} 1 & 0 \end{bmatrix} \left( \begin{bmatrix} s + 2 & -1 \newline 5 & s \end{bmatrix} + \begin{bmatrix} 0 \newline 2 \end{bmatrix} \begin{bmatrix} l_1 & l_2 \end{bmatrix} \right)^{-1} \begin{bmatrix} 0 \newline 2 \end{bmatrix} K_r

\begin{bmatrix} 1 & 0 \end{bmatrix} \left( \begin{bmatrix} s + 2 & -1 \newline 5 & s \end{bmatrix} + \begin{bmatrix} 0 & 0 \newline 2l_1 & 2l_2 \end{bmatrix} \right)^{-1} \begin{bmatrix} 0 \newline 2 \end{bmatrix} K_r

\begin{bmatrix} 1 & 0 \end{bmatrix} \left( \begin{bmatrix} s + 2 & -1 \newline 5 + 2l_1 & s + 2l_2 \end{bmatrix} \right)^{-1} \begin{bmatrix} 0 \newline 2 \end{bmatrix} K_r

As we defined earlier, the characteristic equation is $\det(sI - A + BL_u) = 0$

Which means:

\begin{align*} (s + 2)(s + 2l_2) + (5 + 2l_1) & = 0 \newline s^2 + 2l_2 s + 2s + 4l_2 + 5 + 2l_1 & = 0 \newline s^2 + (2l_2 + 2)s + 4l_2 + 5 + 2l_1 & = 0. \end{align*}

We knew that:

G_{ry}(s) = \dfrac{8}{s^2 + 4s + 8}.

Which must mean that:

\begin{cases} 2l_2 + 2 & = 4 \newline 4l_2 + 5 + 2l_1 & = 8. \end{cases}

We find that $l_1 = -\frac{1}{2}, l_2 = 1$ , which means that:

\boxed{L_u = \begin{bmatrix} -\frac{1}{2} & 1 \end{bmatrix}}

Let’s now find $K_r$ .

K_r = \dfrac{1}{C(-A + BL_u)^{-1} B}.

This is just some basic matrix operations:

K_r = \ldots = \boxed{4}